Another fun problem! It basically relies on the Pythagorean theorem. We usually think of the Pythagorean theorem as referring to the areas of squares on the sides of the triangle, but actually it could be any three similar shapes. In this case, the Pythagorean theorem tells us that the area of semicircle ADB + the area of semicircle BEC = the area of semicircle ABC. ADB + BEC = ABC (all semicircles) Once you see that, the solution is straightforward. Just remove those unshaded parts from both sides of this equation, and we are left with lune ADB + lune BEC = triangle ABC So the green region has the same area as the blue region!
Wow. What a ginormous thought of constructic the puzzle with solution
Another fun problem! It basically relies on the Pythagorean theorem. We usually think of the Pythagorean theorem as referring to the areas of squares on the sides of the triangle, but actually it could be any three similar shapes. In this case, the Pythagorean theorem tells us that the area of semicircle ADB + the area of semicircle BEC = the area of semicircle ABC.
ADB + BEC = ABC (all semicircles)
Once you see that, the solution is straightforward. Just remove those unshaded parts from both sides of this equation, and we are left with
lune ADB + lune BEC = triangle ABC
So the green region has the same area as the blue region!
Thanks for the great insight Scott! That is a very intuitive way to think about this problem.
Well that was fun, thanks for sharing this interesting problem.
My pleasure!
nicely explained